- #1

scothoward

- 29

- 0

Hi,

In the attached file, the question asks to find the three line currents. I understand the solution using Mesh Analysis that was used. However, I am unsure as to why when I convert the Delta Load to a Wye load, I get the wrong answer.

Using Zwye = (1/3)Zdelta, adding the series line impedance with the load impedance, and then finding the line current using a single phase equivalent.

Essentially the Balanced Delta load becomes a Balance Wye load of 4 + j0.666. Then the line impedances will be in series with each individual load. So each impedance should be (4 + j0.666) + (1 + j2). Finally, the line currents will be the phase voltages (100 angle0, 100 angle-120, 100 angle120) divided by the series line and load impedance, 5 + j2.666.

Am I doing something wrong?

In the attached file, the question asks to find the three line currents. I understand the solution using Mesh Analysis that was used. However, I am unsure as to why when I convert the Delta Load to a Wye load, I get the wrong answer.

Using Zwye = (1/3)Zdelta, adding the series line impedance with the load impedance, and then finding the line current using a single phase equivalent.

Essentially the Balanced Delta load becomes a Balance Wye load of 4 + j0.666. Then the line impedances will be in series with each individual load. So each impedance should be (4 + j0.666) + (1 + j2). Finally, the line currents will be the phase voltages (100 angle0, 100 angle-120, 100 angle120) divided by the series line and load impedance, 5 + j2.666.

Am I doing something wrong?